Acceleration Vector in Circular Motion

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mindarson
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Homework Statement



Consider a go-cart moving on a circular track of radius R = 40 m. Suppose it starts from rest and speeds up to 60 km/hr in 20 seconds, with a constant rate of increase of the speed.
Calculate the magnitude of a.



Homework Equations



a(t) = a_t + α

where the first term is the tangential acceleration and the second is the angular acceleration.


The Attempt at a Solution



I begin with the magnitude of a:

mag of a = [(R^2)(α^2) - (R^2)(ω^4)]^.5

Then I calculate ω:

ω = v/R = (60000m/3600s)/40m = .4167 /s

And I can calculate the tangential acceleration:

a_t = Δv/Δt = (60000m/3600s)/20s = .8333 m/s^2

Now I can calculate dω/dt = α:

α = r*a_t = 40 m * .8333 m/s^2 = 33.33 /s^2

Now to calculate the magnitude of the acceleration vector:

mag of a = [(a_t)^2 + α^2]^.5 = 33.34 m/s^2

But apparently this is wrong. Can anyone point out what I've missed? Seems pretty straightforward but I just can't get it.
 
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mindarson said:

Homework Statement



Consider a go-cart moving on a circular track of radius R = 40 m. Suppose it starts from rest and speeds up to 60 km/hr in 20 seconds, with a constant rate of increase of the speed.
Calculate the magnitude of a.

Homework Equations



a(t) = a_t + α

where the first term is the tangential acceleration and the second is the angular acceleration.
[/b]

Did you mean centripetal acceleration instead of angular one?

mindarson said:
Now I can calculate dω/dt = α:
α = r*a_t = 40 m * .8333 m/s^2 = 33.33 /s^2

This is wrong. the tangential acceleration is R times the angular acceleration a_t=αR.ehild
 
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